[LeetCode] 665. Non-decreasing Array
Posted arcsinw
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Description
Given an array with n
integers, your task is to check if it could become non-decreasing by modifying at most 1
element.
We define an array is non-decreasing if array[i] <= array[i + 1]
holds for every i
(1 <= i < n).
Example 1:
Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
Example 2:
Input: [4,2,1]
Output: False
Explanation: You can't get a non-decreasing array by modify at most one element.
Note: The n
belongs to [1, 10,000].
Analyse
给定一个大小为n的数组A
,允许最多修改一个元素的值,判断能否变成一个非递减序列
从左到右遍历数组,当遇到比前一个数A[i-1]
小的数A[i]
的时候,由于可以修改一个元素的值,可以将前一个数A[i-1]
变小,A[i-1] = A[i-2]
,
反映到代码里,直接判断当前数A[i]
是否大于A[i-2]
即可
有一种不满足以上关系的情况(1-2<0)
A[1] < A[0]
比如{4, 2, 3},这种情况就直接忽略4,判断剩下的元素是否非递减的
最终代码如下,这道题我提交了快10次,拿到了8个LeetCode的测试用例,最后看着Solution改自己的版本才AC了
Think more, write less
bool checkPossibility(vector<int>& nums)
{
int last = -100000;
bool state = false;
for (int i = 0; i < nums.size(); i++)
{
if (nums[i] >= last)
{
last = nums[i];
}
else
{
if (state) return false;
if (i==1 || nums[i] >= nums[i-2])
{
last = nums[i];
}
state = true;
}
}
return true;
}
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